Symmetry group of a two-dimensional isotropic harmonic oscillator

Question

The Lagrangian and the Hamiltonian of a two-dimensional isotropic oscillator (with $m=\omega=1$) are $$L=\frac{1}{2}(v^2_1+v_2^2-q_1^2-q_2^2)\tag{1}$$ and $$H=\frac{1}{2}(p^2_1+p_2^2+q_1^2+q_2^2),\tag{2}$$ respectively. The expression for the Lagrangian suggests that it has $O(2,2)$ symmetry and that of the Hamiltonian suggests that it has $O(4)$ symmetry. Can someone explain which is correct and why?

Answer 1

First of all, in this answer we assume that OP's 2 symmetries refer to (possibly velocity dependent) quasi-symmetries of a Lagrangian and a Hamiltonian action, respectively, so that Noether's theorem applies.

1. OP is right that the quadratic form $L(q,v)$ of signature (2,2) is invariant under the indefinite orthogonal group $O(2,2)$. But the $O(2,2)$ invariance turns out to be neither sufficient nor necessary$^1$.

   Concerning not sufficient: Recall that in the Lagrangian action, the velocity configuration $t\mapsto\dot{q}(t)$ is not independent of the position configuration $t\mapsto q(t)$, cf. e.g. my Phys.SE answer here.

   In fact, since time $t$ is not varied $\delta t=0$ in the isotropic HO, the infinitesimal symmetry variation of velocity $\delta \dot{q}$ is completely determined by the infinitesimal symmetry variation of position $\delta q= \epsilon f(q,\dot{q})$.

2. OP is right that the quadratic form $H(q,p)$ of signature (4,0) is invariant under the orthogonal group $O(4)$. However, in the Hamiltonian formulation, the phase space transformation should respect the symplectic structure, so the quasi-symmetry group is a subgroup of $O(4)$.

It turns out that in both cases the corresponding quasi-symmetry group is the same, namely $U(2)$, cf. this related Phys.SE post.

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$^1$ Concerning not necessary: The pertinent quasi-symmetry group $U(2)$ [which has 4 compact dimensions] is not a subgroup of $O(2,2)$ [which has only a maximal compact subgroup $O(2)\times O(2)$].